Problem: Let $h(x)=-3x^4+8x^3+18x^2$. What is the absolute maximum value of $h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $270$ (Choice B) B $0$ (Choice C) C $135$ (Choice D) D $h$ has no maximum value
Solution: Let's first find the relative extremum points of $h$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=-12x(x-3)(x+1)$. $h'(x)=0$ for $x=-1,0,3$. $h'$ is defined for all real numbers. Therefore, our critical points are $x=-1$, $x=0$, and $x=3$. Our critical points divide the function's domain (which is all real numbers) into four intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $x< \llap{-}1$ $\llap{-}1<x<0$ $0<x<3$ $x>3$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-1$ $x=-2$ $h'\left(-2\right)=120>0$ $h$ is increasing $\nearrow$ $-1<x<0$ $x=-\dfrac12$ $h'\left(-\dfrac{1}{2}\right)=-\dfrac{21}{2}<0$ $h$ is decreasing $\searrow$ $0<x<3$ $x=1$ $h'\left(1\right)=48>0$ $h$ is increasing $\nearrow$ $x>3$ $x=4$ $h'\left(4\right)=-240<0$ $h$ is decreasing $\searrow$ Now let's look at all the critical points: $x$ $h(x)$ Before After Verdict $-1$ $7$ $\nearrow$ $\searrow$ Maximum $0$ $0$ $\searrow$ $\nearrow$ Minimum $3$ $135$ $\nearrow$ $\searrow$ Maximum Let's imagine ourselves walking on the graph of $h$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up until $(-1,7)$, then go down until $(0,0)$, then up again until $(3,135)$, and then forever go down. This means that $\lim_{x\to-\infty}h(x)=\lim_{x\to +\infty}h(x)=-\infty$, which means $h$ has no minimum value. However, $h$ does reach an absolute maximum point at $(3,135)$, which means its absolute maximum value is $135$. In conclusion, the absolute maximum value of $h$ is $135$.